# Show that this sequence is positive with probability approaching one

Let $$(X_n)$$ be a sequence of square integrable real random variables on a probability space $$(\Omega,\mathcal F,P)$$. Suppose that

$$E[X_n\mid \mathcal A]\to 1_A \quad P\text{-a.s.}$$

$$V[X_n\mid \mathcal A]\to 0 \quad P\text{-a.s.}$$

as $$n\to \infty$$, where $$\mathcal A\subset \mathcal F$$ is a sub $$\sigma$$-algebra, and $$A\in\mathcal A$$ has positive $$P$$-measure. Can I show then that

$$P[X_n\geq 0]\to 1 \quad \text{as } n\to\infty \quad?$$

Any help is very appreciated.

• Are you sure that both those conditions can even happen at the same time if $0<P(A)<1$?
– Ian
Jun 17, 2021 at 14:08
• @Ian If $X_n=1_A$ for all $n$ with $A\in\mathcal A$ then I believe both conditions hold no? Jun 17, 2021 at 14:12
• Never mind, that's right.
– Ian
Jun 17, 2021 at 14:12

No, I don't think we can conclude $$P(X_n \ge 0) \rightarrow 1$$. Let $$\Omega = [0,1]$$ with $$P$$ the Lebesgue measure and $$\mathcal A$$ the sigma-algebra on $$[0,1]$$ generated by the Borel subsets of $$[0,\frac 12]$$. Let $$A = [0,\frac 12]$$ and \begin{align*} X_n(x) := \begin{cases} 1 & x \in A \\ \frac 1n & x \in (\frac 12, \frac 34] \\ -\frac 1n & x \in (\frac 34,1] \end{cases}. \end{align*} Then $$\mathbb{E}[X_n | \mathcal A] = 1_A$$ for all $$n$$ so clearly $$\mathbb{E}[X_n | \mathcal A] \rightarrow 1_A$$, and $$V(X_n | \mathcal A) = \frac{1}{n^2}1_{(\frac 12, 1]} \rightarrow 0$$ almost surely. However, we have $$P(X_n < 0) = \frac 14$$ for all $$n$$, so $$P(X_n \ge 0) \not \rightarrow 1$$.

• Typo: the $0$ at the end should be $1$. Anyway, I suppose an obvious followup is "can you ensure $P(X_n>-\epsilon) \to 1$ for each $\epsilon>0$?"
– Ian
Jun 17, 2021 at 16:41
• Isn't $V(X_n | \mathcal A) = 0$ $P$-a.s. for all $n$? Jun 21, 2021 at 14:03
• @Alphie I think I did make a mistake in $V(X_n|\mathcal A)$, but I don't think it's $0$ a.s. We have $(X_n - \mathbb{E}[X_n|\mathcal A])^2 = (X_n - 1_A)^2 = \frac{1}{n^2} 1_{(\frac 12,1]}$, so $V(X_n|\mathcal A) = \mathbb{E}[(X_n - \mathbb{E}[X_n|\mathcal A])^2|\mathcal A] = \frac{1}{n^2} 1_{(\frac 12,1]}$ which still converges to $0$. Jun 21, 2021 at 15:56
• It seems to me that $\mathbb E[1_{(\frac 12,1]} | \mathcal A]=0$ a.s. no? Jun 21, 2021 at 16:41
• @Alphie No, $(\frac 12,1]$ is measurable with respect to $\mathcal A$ because it is $A^c$, and $A$ is $\mathcal A$-measurable. Jun 21, 2021 at 16:49

In general no: let $$X_n=\mathbf{1}_A+Y_n$$, where $$Y_n$$ is independent of $$\mathcal A$$. The first assumption translates as $$\mathbb E\left[Y_n\right]\to 0$$ and the second one as $$\operatorname{Var}(Y_n)\to 0$$.

Moreover, $$\mathbb P(X_n\geqslant 0)=\mathbb P(A\cap \{1+Y_n\geqslant 0)+\mathbb P(A^c\cap \{Y_n\geqslant 0)$$ and by independence, $$\mathbb P(X_n\geqslant 0)=\mathbb P(A)\mathbb P(1+Y_n\geqslant 0)+\mathbb P(A^c)\mathbb P(Y_n\geqslant 0).$$ Let $$(Y_n)_{n\geqslant 1}$$ be a sequence of independent random variables such that for each $$n$$, $$Y_n$$ taked the values $$1/n$$ and $$-1/n$$ with probability $$1/2$$. Letting $$\mathcal A=\sigma(Y_1)$$ and $$A=\{Y_1=1\}$$, the assumptions on fulfills the assumptions and for $$n\geqslant 2$$, $$\mathbb P(X_n\geqslant 0)=\mathbb P(A\cap \{1+X_n\geqslant 0\})+\mathbb P(A^c\cap \{X_n\geqslant 0\})= \mathbb P(A)+\mathbb P(A^c)/2<1,$$ as $$\mathbb P(1+Y_n\geqslant 0)=1$$ and $$\mathbb P(Y_n\geqslant 0)=1/2$$.

• Thank you for your answer. Two questions: 1) Isn't $\mathbb P(X_n\geqslant 0)= \mathbb P(A)+\mathbb P(A^c)/2$? 2) How do you ensure that $Y_n$ is independent from $\mathcal A$? Jun 21, 2021 at 13:45
• @Alphie Ii have added more details. Actually it seems that $\mathbb P(X_n\geqslant 0)= 1/2$. Jun 21, 2021 at 15:12
• Thank you it is clearer now. I would say that $\mathbb P(1+Y_n\geqslant 0)=1$ so that $\mathbb P(X_n\geqslant 0)= 3/4$. Jun 21, 2021 at 15:38
• Oh you are right, I edited again. Jun 21, 2021 at 15:41